What is the answer to x^2-9i=0 ?

The answer to x^2-9i=0 is x= 3/(sqrt(2))+3/(sqrt(2))i,x=-3/(sqrt(2))-3/(sqrt(2))i

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x^2-9i=0

x^{2} - 9 i = 0

x = \frac{3}{2} + \frac{3}{2} i, x = - \frac{3}{2} - \frac{3}{2} i

Solution steps

x^{2} - 9 i = 0

Substitute

x = a + bi

(a + bi)^{2} - 9 i = 0

Expand

(a + bi)^{2} - 9 i : (a^{2} - b^{2}) + i (- 9 + 2 ab)

(a^{2} - b^{2}) + i (- 9 + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2}) + i (- 9 + 2 ab) = 0 + 0 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[a^{2} - b^{2} = 0 - 9 + 2 ab = 0]

[a^{2} - b^{2} = 0 - 9 + 2 ab = 0] : (a = \frac{3}{2}, a = - \frac{3}{2}, b = \frac{3}{2} b = - \frac{3}{2})

Substitute back

x = a + bi

x = \frac{3}{2} + \frac{3}{2} i, x = - \frac{3}{2} - \frac{3}{2} i

(x + 2 - 1) (x - 2 i) = 0

(x - i y)^{2} = 5 - 12 i

(p + 3 i) \cdot (4 + q i) = 15 + 9 i

(x + 4 i) (7 + y i) = 23 + 43 i

\frac{1}{2} - (y - 4) i = (x + 1) + 2 i

What is the answer to x^2-9i=0 ?
The answer to x^2-9i=0 is x= 3/(sqrt(2))+3/(sqrt(2))i,x=-3/(sqrt(2))-3/(sqrt(2))i